Question

100 mL of a ${10}^{-2}M$ weak monoacidic base $(Kb=\frac{1}{3}\times {10}^{-11}$ at $25{}^{\circ }C)$ is titrated with $5\times {10}^{-3}M$ $HCl$ in water at $25{}^{\circ }C$. The pH of the solution at equivalence point is: $(Kw=1\times {10}^{-14}$ at $25{}^{\circ }C)$
(take log2 = 0.3)

• A. 11.3
• B. 2.5
• C. 2.7
• D. 11.5

Answer 1

The correct option is C 2.7

Let the base be $BOH$.
So in mmols $=100\times {10}^{-2}=1$ mmol of $BOH$
Sine the base is monoacidic and $HCl$ is monobasic acid, so for complete titration,
$5\times {10}^{-3}\times V=1$ $\Rightarrow V=200ml$
After neutralization, salt will be formed and its concentration will be, $\left[BCl\right]=\frac{1}{100+200}=\frac{1}{3}\times {10}^{-2}M$
Now from hydrolysis of salt,
$B^{+}+H_{2}O\stackrel{K_h=\frac{K_w}{K_b}=3\times {10}^{-3}}{\rightleftharpoons }BOH+H^{+}$
$\frac{1}{3}\times {10}^{-2}-xxx$
$K_h=3\times {10}^{-3}=\frac{x^2}{\frac{1}{3}\times {10}^{-2}-x}\Rightarrow x=2\times {10}^{-3}M$
So, $\left[H^{+}\right]=2\times {10}^{-3}M\Rightarrow pH=2.7$